Let $f$ Lebesgue integrable function.Then, is there measurable function $h$ such that $h$ is increasing, $\lim _{x\to \infty } h(x)=\infty $, and $fh$ is integrable ?
I guess when $n$ is large, $h_n(x):=\log{\log {\cdots \log{x}}}=\log^n{x}$ satisfies. But I cannot prove it.
On
Assuming that all these functions are on $\mathbb{R}^d$, $h$ cannot be lebesgue integrable if it goes to $\infty$ as we go far away from the origin. If $f$ is integrable, we know that $\int_{B^c}|f|<\epsilon$ for a sufficiently large ball $B$ around the origin. Take a sequence of such increasing balls $\{B_n\}$ such that the integral of $|f|$ over $B_n-B_{n-1}$ is less than $\epsilon/2^n$. Now, build $h$ such that $h$ is equal to $2^{n/2}$ in the region between $B_{n-1}$ and $B_n$. Define $h$ to be $0$ inside $B_1$. This function $h$ is increasing and goes to infinity as we go away from the origin. Also, $fh$ is integrable as the sum $\sum \frac{\epsilon}{2^{n/2}}$ is finite.
If all these functions are defined on $\mathbb{R}$ and we want $h$ to be increasing in the usual sense, then we can just take $h'$ instead of $h$, where $h'=h$ for the positive real line, and $h'=-h$ for the negative real line. $h'$ is now increasing and satisfies the other conditions.
One can write down such a function $h$ explicitly in terms of $f$. First assume that $|f|>0$. define $h(x)=\frac 1 {\sqrt {\int_x^{\infty} |f|}}$ if $x \geq 1$ and $0$ otherwise. Clearly. $h(x) \to \infty$ as $x \to \infty$. Now let $g(x)= {\int_x^{\infty} |f|}$. Then $\int |f|h =\int_0^{\infty} \frac {-g'(x)} {\sqrt {{g(x)}}}=2\sqrt {g(1)}$ so $|f|h$ is integrable. For the general case replace $f$ by $|f|+e^{-|x|}$. The function $h$ which works for this also works for $f$.